How to solve nonlinear coupled ode by Shooting method .
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Respected sir ,
I am getting problem to solve non linear coupled BVP by shooting method . Can you help me to solve that problem?
My problem is
F'''=(1/(1+epsilon1*(1-G)))*(A*F'+.5*t*A*F''+(F')^2-F*F''+epsilon1*G'*F''-lambda*G-delta*H+M*F'-(-1+epsilon1*G-epsilon1)*bita*F');
G''=(1/(1+epsilon2*G+Nr))*(Pr*(2*A*G+.5*A*t*G'+F'*G-F*G'-M*Ec*(F')^2-Ec*(1-epsilon1*G'+epsilon1)*(F'')^2)-epsilon2*(G')^2);
H''= Sc*(2*A*H+.5*A*t*H'+F'*H-F*H'+Rex*Zai*H);
where A=0 ; epsilon1= 0 ; epsilon2=1; lambda= 1; delta=1; bita=0; Nr=.1; Pr=5; M=.5;
Ec=.1 ; Sc=1; Rex= .3 ; Zai= .1 ;
and F(0)=0 , F'(0)=1 F'(infity)=0 G(0)=1 G(infity)=0 H(0)= 1 H(infity)=0
and F''(0) , G'(0) & H'(0) we have to guess
so tell me how to solve by shooting method with using rk -4 method .
2 comentarios
darova
el 13 de Oct. de 2019
What have you tried? What ideas do you have? Did you see examples in MATLAB help?
Chandan Kumawat
el 14 de Oct. de 2019
Respuesta aceptada
Más respuestas (1)
darova
el 14 de Oct. de 2019
Try bvp4c
Suggestion:
F0 = y(1);
%% ...
H1 = y(7);
% and use these variables to make your code more redable
dy(1) = F1;
%% ..
You can also use temporary variables to make your code simpler
dy(3) = 1/(1+e1*(1-G0))* ...
(A*F1 + 0.5*t*A*F2 + F1^2 - F0*F2 + e1*G1*F2 - lambda*G0 - delta*H0 + M*F1 - (-1+e1*G0-e1)*bita*F1);
%%
TEMP0 = 1/(1+e1*(1-G0));
TEMP1 = 0.5*t*A*F2;
TEMP2 = e1*G1*F2;
TEMP3 = (-1+e1*G0-e1)*bita*F1;
dy(3) = TEMP0 * (A*F1 + TEMP1 + F1^2 - F0*F2 + TEMP2 - lambda*G0 - delta*H0 + M*F1 - TEMP3);
See attached scripts
10 comentarios
Chandan Kumawat
el 14 de Oct. de 2019
darova
el 14 de Oct. de 2019
How do you think those points mean?

Chandan Kumawat
el 15 de Oct. de 2019
darova
el 15 de Oct. de 2019
Please see the picture

Chandan Kumawat
el 15 de Oct. de 2019
darova
el 15 de Oct. de 2019
Impossible. What is Untitled6 ? Why don't you run the script i attached Untitled21?
Chandan Kumawat
el 15 de Oct. de 2019
darova
el 15 de Oct. de 2019
constant value of M Ec Pr
Are you sure those values are correct? Maybe for something values diving by zero occurs?
Chandan Kumawat
el 17 de Oct. de 2019
darova
el 17 de Oct. de 2019
I just changed main code
init = [0 1 -0.58 1 -1.52 1 -1.12];
% solinit = bvpinit([0 2],zeros(1,7));
% sol = bvp4c(@new,@bvpf,solinit);
[t,y] = ode45(@new,[0 2], init);
% plot(sol.x,sol.y)
plot(t,y)
legend('F','dF','d2F','G','dG','H','dH')
I took initial conditions from last calc. There is no rule for F2(0) G1(0) & H1(0) values, only guessing or something like bvp4c

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